Beans
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4451 Accepted Submission(s): 2103
Problem Description
Bean-eating
is an interesting game, everyone owns an M*N matrix, which is filled
with different qualities beans. Meantime, there is only one bean in any
1*1 grid. Now you want to eat the beans and collect the qualities, but
everyone must obey by the following rules: if you eat the bean at the
coordinate(x, y), you can’t eat the beans anyway at the coordinates
listed (if exiting): (x, y-1), (x, y+1), and the both rows whose
abscissas are x-1 and x+1.
is an interesting game, everyone owns an M*N matrix, which is filled
with different qualities beans. Meantime, there is only one bean in any
1*1 grid. Now you want to eat the beans and collect the qualities, but
everyone must obey by the following rules: if you eat the bean at the
coordinate(x, y), you can’t eat the beans anyway at the coordinates
listed (if exiting): (x, y-1), (x, y+1), and the both rows whose
abscissas are x-1 and x+1.
Now, how much qualities can you eat and then get ?
Input
There
are a few cases. In each case, there are two integer M (row number) and
N (column number). The next M lines each contain N integers,
representing the qualities of the beans. We can make sure that the
quality of bean isn't beyond 1000, and 1<=M*N<=200000.
are a few cases. In each case, there are two integer M (row number) and
N (column number). The next M lines each contain N integers,
representing the qualities of the beans. We can make sure that the
quality of bean isn't beyond 1000, and 1<=M*N<=200000.
Output
For each case, you just output the MAX qualities you can eat and then get.
Sample Input
4 6
11 0 7 5 13 9
78 4 81 6 22 4
1 40 9 34 16 10
11 22 0 33 39 6
Sample Output
242
题意：
每一个格子有一个豆子，豆子有质量，每一行相邻的两个豆子不能同时选必须相隔，每一列也不能同时选，选了一列的某几个豆子则它的上一列和下一列不能选，问最多选到的豆子质量。
代码：
/*
两次dp,算出每一行的最大值，再用每一行的最大值组成一列算出这列的最大值即可。
*/
#include<iostream>
#include<string>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<queue>
#include<stack>
using namespace std;
int n,m;
int x[];
int dp[][];
int a[];
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(x,,sizeof(x));
for(int i=;i<=n;i++)
{
memset(dp,,sizeof(dp));
for(int j=;j<=m;j++)
{
scanf("%d",&a[j]);
}
dp[][]=a[];
dp[][]=;
for(int j=;j<=m;j++)
{
dp[j][]=dp[j-][]+a[j];
dp[j][]=max(dp[j-][],dp[j-][]);
}
x[i]=max(dp[m][],dp[m][]);
}
memset(dp,,sizeof(dp));
dp[][]=x[];
dp[][]=;
for(int i=;i<=n;i++)
{
dp[i][]=dp[i-][]+x[i];
dp[i][]=max(dp[i-][],dp[i-][]);
}
int sum=max(dp[n][],dp[n][]);
printf("%d\n",sum);
}
return ;
}